C.R.Krishna Prasad, BIT Bangalore-4 Operations research

C.R.Krishna Prasad, BIT Bangalore-4

Operations research


Origin and Development

The main origin of OR was during the II world war Military management in England called upon a team

of scientists to study the Strategic* and Tactical* problems related to air and land defence of the country.

Since they were having very limited resources, it was necesent osary to decide upon the most effective utilization of them., eg the efficient ocean transport, effective bombing, etc.

(* premeditated)


Contd….

Their mission was to formulate the specific proposals and plans for aiding the military commands to arrive at the decisions on optimal utilization of scarce military resources and efforts and also to implement the decision effectively.

Scientific and systematic approaches involved in OR provided a good intellectual support to the strategic initiatives of the military commands.


i

One group in britain came to known as Blacket circus.(radar OR Unit in gun site).

The US military team named operational analysis ,operational evaluation, operational research, system analysis ,system evaluation, system research & management.

But military team were dealing with research on (military) operation, the work of this team

of scientists named as Operation Research in England.


Introduction

End of the war, the success of military teams attracted the attention of industrial managers Who were seeking their complex executive type problems.

The most common problem was : what method should be adopted so that the total cost is minimum or total profit is maximum.


The first mathematical technique in this field (called the simplex method of linear programming) was developed by american mathematician,George B Damtzig.

Since then techniques & application have been developed through the effort and cooperation of interested individual in academic institutions and industry both.


Definitions of OR

OR is the systematic method oriented study of the basic structure, characteristic, functions and relationships of an organisation to provide the executive with a sound, scientific and quantitative basis for decision making.

OR is concerned with scientifically deciding how to best design and operate man-machine systems usually requiring the allocation of resources.


Contd..

OR is the art of giving bad answers to the

problems to which otherwise worse answers

Are given. An art of winning the war without actually

fighting it.


Characteristics of OR OR approaches problem solving and decision

making from a total system’s perspective It is interdisciplinary model Model building and mathematical manipulation

provide the methodology . OR is for operations economy Primary focus on decision making.


PHASES OF OPERATIONS RESEARCH STUDY

I Formulation The Problem: What are the objectives, controlled variables,

uncontrolled variable constraints

IIConstructing a mathematical method

A mathematical model should include a) decision variable and parameter,,objective functions&constraints.


III Deriving the solutions from the model OR models include LPP, Transportation, Assignment,

Queuing models, Network analysis, Job sequencing, Replacement models, simulation models

IV Testing the model and solution(updating the model)

V: Controlling the solution VI: Implementing the solution


Techniques of OR

Distribution/allocation Models Waiting line/queuing models Production/inventory model Competitive strategy model/games

theory Network analysis Job sequencing models


Replacement models Markovian models Simulation models


Scope of O.R.

In agriculture In Finance- BEP,Capital budgeting,

SAPM, cash flow,financial planning In Industry In Marketing In Personnel management In Production management In Life Insurance


Limitations Magnitude of computation Absence of quantification Distance between manager and OR

experts


Linear Programming Problems


- Formulation

Linear Programming is a mathematical technique for optimum

allocation of limited or scarce resources, such as labour, material,

machine, money, energy and so on , to several competing activities

such as products, services, jobs and so on, on the basis of a given

criteria of optimality.


Contd….

The term ‘Linear’ is used to describe the proportionate relationship of

two or more variables in a model. The given change in one variable

will always cause a resulting proportional change in another variable.

The word , ‘Programming’ is used to specify a sort of planning that

involves the economic allocation of limited resources by adopting a

particular course of action or strategy among various alternatives

strategies to achieve the desired objective.


Structure of Linear Programming model.

The general structure of the Linear Programming model

essentially consists

of three components.

i) The activities (Decision variables) and their

relationships

ii) The objective function and

iii) The constraints


i) The activities are represented by X1, X2, X3 ……..Xn. These are known as Decision variables.

ii) The objective function of an LPP (Linear Programming Problem) is a mathematical representation of the objective in terms a measurable quantity such as profit, cost, revenue, etc.Optimize (Maximize or Minimize) Z=C1X1 +C2X2+ ………..Cn Xn

Where Z is the measure of performance variable X1, X2, X3, X4…..Xn are the decision variablesAnd C1, C2, …Cn are the parameters that give contribution to decision variables.

iii) Constraints are the set of linear inequalities and/or equalities which impose restriction of the limited resources


General Mathematical Model of an LPP

Optimize (Maximize or Minimize) Z=C1 X1 + C2 X2 +……+CnXn

Subject to constraints,

a11X1+ a 12X2+………………+ a 1nXn (<,=,>) b1

a21X1+ a 22X2+………………+ a 2nXn (<,=,>) b2

a31X1+ a 32X2+………………+ a 3nXn (<,=,>) b3

am1X1+ a m2X2+………………+ a mnXn (<,=,>) bm

and X1, X2 ….Xn > 0


Guidelines for formulating Linear Programming model

i) Identify and define the decision variable of the problem

ii) Define the objective function

iii) State the constraints to which the objective function should be

optimized (i.e. either Maximization or Minimization)

iv) Add the non-negative constraints from the consideration that the

negative values of the decision variables do not have any valid physical

interpretation


Example 1.

A firm is engaged in producing two products. A and B. Each unit of

product A requires 2 kg of raw material and 4 labour hours for

processing, where as each unit of B requires 3 kg of raw materials

and 3 labour hours for the same type. Every week, the firm has an

availability of 60 kg of raw material and 96 labour hours. One unit of

product A sold yields Rs.40 and one unit of product B sold gives

Rs.35 as profit.

Formulate this as an Linear Programming Problem to determine as

to how many units of each of the products should be produced per

week so that the firm can earn maximum profit.



Let X1 and X2 be the number of units of product A and product

B produced per week.


Since the profits of both the products are given,

the objective function is to maximize the profit.

MaxZ = 40X1 + 35X2



optimized (i.e. Maximization or Minimization)

There are two constraints one is raw material constraint and the other

one is labour constraint..

The raw material constraint is given by

2X1 + 3X2 < 60

The labour hours constraint is given by

4X1 + 3X2 < 96


Finally we have,

MaxZ = 40X1 + 35X2


2X1 + 3X2 < 60

4X1 + 3X2 < 96

X1,X2 > 0


Example 2.

The agricultural research institute suggested the farmer to spread out

atleast 4800 kg of special phosphate fertilizer and not less than 7200 kg of

a special nitrogen fertilizer to raise the productivity of crops in his fields.

There are two sources for obtaining these – mixtures A and mixtures B.

Both of these are available in bags weighing 100kg each and they cost

Rs.40 and Rs.24 respectively. Mixture A contains phosphate and nitrogen

equivalent of 20kg and 80 kg respectively, while mixture B contains these

ingredients equivalent of 50 kg each. Write this as an LPP and determine

how many bags of each type the farmer should buy in order to obtain the

required fertilizer at minimum cost.



Let X1 and X2 be the number of bags of mixture A and mixture B.


The cost of mixture A and mixture B are given ;

the objective function is to minimize the cost

Min.Z = 40X1 + 24X2



optimized.

The above objective function is subjected to following constraints.

20X1 + 50X2 >4800 Phosphate requirement

80X1 + 50X2 >7200 Nitrogen requirement

X1, X2 >0


Finally we have,

Min.Z = 40X1 + 24X2

is subjected to three constraints

20X1 + 50X2 >4800

80X1 + 50X2 >7200

X1, X2 >0


Example 3.

A manufacturer produces two types of models M1 and M2.Each

model of the type M1 requires 4 hours of grinding and 2 hours of

polishing; where as each model of M2 requires 2 hours of grinding

and 5 hours of polishing. The manufacturer has 2 grinders and 3

polishers. Each grinder works for 40 hours a week and each

polisher works 60 hours a week. Profit on M1 model is Rs.3.00 and

on model M2 is Rs.4.00.Whatever produced in a week is sold in the

market. How should the manufacturer allocate his production

capacity to the two types of models, so that he makes maximum

profit in a week?



Let X1 and X2 be the number of units of M1 and M2 model.


Since the profits on both the models are given, the objective function

is to maximize the profit.

Max Z = 3X1 + 4X2



optimized (i.e. Maximization or Minimization)

There are two constraints one for grinding and the other for polishing.

The grinding constraint is given by

4X1 + 2X2 < 80

No of hours available on grinding machine per week is 40 hrs. There

are two grinders. Hence the total grinding hour available is 40 X 2 = 80

hours.


The polishing constraint is given by

2X1 + 5X2 < 180

No of hours available on polishing machine per week is 60

hrs. There are three grinders. Hence the total grinding hour

available is 60 X 3 = 180 hours.


Finally we have,

Max Z = 3X1 + 4X2


4X1 + 2X2 < 80

2X1 + 5X2 < 180

X1,X2 > 0


Example 4.

A firm can produce 3 types of cloth, A , B and C.3 kinds of wool

are required Red, Green and Blue.1 unit of length of type A

cloth needs 2 meters of red wool and 3 meters of blue wool.1

unit of length of type B cloth needs 3 meters of red wool, 2

meters of green wool and 2 meters of blue wool.1 unit type of C

cloth needs 5 meters of green wool and 4 meters of blue wool.

The firm has a stock of 8 meters of red,10 meters of green and

15 meters of blue. It is assumed that the income obtained from

1 unit of type A is Rs.3, from B is Rs.5 and from C is

Rs.4.Formulate this as an LPP.( December2005/January 2006)



Let X1, X2 and X3 are the quantity produced of cloth type A,B and C

respectively.


The incomes obtained for all the three types of cloths are given;

the objective function is to maximize the income.

Max Z = 3X1 + 5X2 + 4X3



optimized.

The above objective function is subjected to following three

constraints.

2X1 + 3X2 < 8

2X2 + 5X3 < 10

3X1 + 2X2 + 4X3 < 15

X1, X2 X3 >0


Finally we have,

Max Z = 3X1 + 5X2 + 4X3


2X1 + 3X2 < 8

2X2 + 5X3 < 10

3X1 + 2X2 + 4X3 < 15

X1, X2 X3 >0


Example 5.

A Retired person wants to invest upto an amount of Rs.30,000 in fixed

income securities. His broker recommends investing in two Bonds: Bond

A yielding 7% and Bond B yielding 10%. After some consideration, he

decides to invest atmost of Rs.12,000 in bond B and atleast Rs.6,000 in

Bond A. He also wants the amount invested in Bond A to be atleast equal

to the amount invested in Bond B. What should the broker recommend if

the investor wants to maximize his return on investment? Solve

graphically. (January/February 2004)



Let X1 and X2 be the amount invested in Bonds A and B.


Yielding for investment from two Bonds are given; the

objective function is to maximize the yielding.

Max Z = 0.07X1 + 0.1X2


iii) State the constraints to which the objective function

should be optimized.


constraints.

X1 + X2 < 30,000

X1 > 6,000

X2 < 12,000

X1 -- X2 >0

X1, X2 >0


Finally we have,

MaxZ = 0.07X1 + 0.1X2


X1 + X2 < 30,000

X1 > 6,000

X2 < 12,000

X1 -- X2 >0

X1, X2 >0


Minimization problems

Example 6.

A person requires 10, 12, and 12 units chemicals A, B and C

respectively for his garden. A liquid product contains 5, 2 and 1

units of A,B and C respectively per jar. A dry product contains 1,2

and 4 units of A,B and C per carton.

If the liquid product sells for Rs.3 per jar and the dry product sells

for Rs.2 per carton, how many of each should be purchased, in

order to minimize the cost and meet the requirements?



Let X1 and X2 be the number of units of liquid and dry products.


The cost of Liquid and Dry products are given ;

The objective function is to minimize the cost

Min. Z = 3X1 + 2X2



optimized.


constraints.

5X1 + X2 >10

2X1 + 2X2 >12

X1 + 4X2 >12

X1, X2 >0


Finally we have,

Min. Z = 3X1 + 2X2


5X1 + X2 >10

2X1 + 2X2 >12

X1 + 4X2 >12

X1, X2 >0


Example 7.

A Scrap metal dealer has received a bulk order from a customer for a

supply of atleast 2000 kg of scrap metal. The consumer has specified

that atleast 1000 kgs of the order must be high quality copper that can

be melted easily and can be used to produce tubes. Further, the

customer has specified that the order should not contain more than

200 kgs of scrap which are unfit for commercial purposes. The scrap

metal dealer purchases the scrap from two different sources in an

unlimited quantity with the following percentages (by weight) of high

quality of copper and unfit scrap.


The cost of metal purchased from source A and source B are Rs.12.50

and Rs.14.50 per kg respectively. Determine the optimum quantities of

metal to be purchased from the two sources by the metal scrap dealer

so as to minimize the total cost (February 2002)

Source A Source B

Copper 40% 75%

Unfit Scrap 7.5% 10%



Let X1 and X2 be the quantities of metal to be purchased from the two

sources A and B.


The cost of metal to be purchased by the metal scrap dealer are given;

the objective function is to minimize the cost

Min. Z = 12.5X1 + 14.5X2



optimized.

The above objective function is subjected to following three constraints.

X1 + X2 >2,000

0.4X1 + 0.75X2 >1,000

0.075X1 + 0.1X2 + 4X3 < 200

X1, X2 >0


Finally we have,

Min. Z = 12.5X1 + 14.5X2


X1 + X2 >2,000

0.4X1 + 0.75X2 >1,000

0.075X1 + 0.1X2 + 4X3 < 200

X1, X2 >0


Example 8.A farmer has a 100 acre farm. He can sell all tomatoes, lettuce or

radishes and can raise the price to obtain Rs.1.00 per kg. for

tomatoes , Rs.0.75 a head for lettuce and Rs.2.00 per kg for

radishes. The average yield per acre is 2000kg.of tomatoes, 3000

heads of lettuce and 1000 kgs of radishes. Fertilizers are available

at Rs.0.50 per kg and the amount required per acre is 100 kgs for

each tomatoes and lettuce and 50kgs for radishes. Labour

required for sowing, cultivating and harvesting per acre is 5 man-

days for tomatoes and radishes and 6 man-days for lettuce. A total

of 400 man-days of labour are available at Rs.20.00 per man-day.

Formulate this problem as LP model to maximize the farmers profit.



Let X1 and X2 and X3 be number acres the farmer grows

tomatoes, lettuce and radishes respectively.


The objective of the given problem is to maximize the profit.

The profit can be calculated by subtracting total expenditure

from the total sales

Profit = Total sales – Total expenditure


The farmer produces 2000X1 kgs of tomatoes, 3000X2

heads of lettuce, 1000X3 kgs of radishes.

Therefore the total sales of the farmer will be

= Rs. (1 x 2000X1 + 0.75 x 3000X2 + 2 x 100X3)

Total expenditure (fertilizer expenditure) will be

= Rs.20 ( 5X1 + 6X2 + 5X3 )

Farmer’s profit will be

Z = (1 x 2000X1 + 0.75 x 3000X2 + 2 x 100X3) –

{ [0.5 x 100 x X1+0.5 x 100 x X2 + 50xX3]+ [20 x 5 x X1+20

x 6 x X2 + 20 x 5 x X3]}

=1850X1 + 2080X2 + 1875X3


Therefore the objective function is

Maximise Z = 1850X1 + 2080X2 + 1875X3


optimized.

The above objective function is subjected to following constraints.

Since the total area of the firm is 100 acres

X1 + X2 + X3 < 100

The total man-days labour is 400 man-days

5X1 + 6X2 + 5X3 < 400


Finally we have,

Maximise Z = 1850X1 + 2080X2 + 1875X3


X1 + X2 + X3 < 100

5X1 + 6X2 + 5X3 < 400

X1, X2 X3 >0


Assumptions of Linear Programming

Certainty.

In all LP models it is assumed that, all the model parameters such as

availability of resources, profit (or cost) contribution of a unit of decision

variable and consumption of resources by a unit of decision variable

must be known with certainty and constant.

Divisibility (Continuity)

The solution values of decision variables and resources are assumed to

have either whole numbers (integers) or mixed numbers (integer (digit)or

fractional (partial/part)). However, if only integer variables are desired,

then Integer programming method may be employed.


Additivity

The value of the objective function for the given value of decision

variables and the total sum of resources used, must be equal to the

sum of the contributions (Profit or Cost) earned from each decision

variable and sum of the resources used by each decision variable

respectively. /The objective function is the direct sum of the

individual contributions of the different variables

Linearity

All relationships in the LP model (i.e. in both objective function and

constraints) must be linear.


Problem : A company has four factories F1,F2,F3,F4 and manufacturing the same product. Production and raw material costs differ from factory to factory and are given in the following table in the first two rows. The transportation costs are from the factories to sales depots, S1,S2,S3,and S4are also given. The last two columns in the table give the sales price and the total requirement at each depot. The production capacity of each factory is given in the last row.


F1 F2 F3 F4

Production cost/unit 15 18 14 14

Raw material cost/unit 10 9 12 9


Determine the most profitable production and distribution schedule and the corresponding profit. The surplus production should be taken to yield zero profit.

Transp.cost S1

3 9 5 4 Selling price.

34

Requirement 80

S2 1 7 4 5 32 120

S3 5 8 3 6 31 150

10 150 50 100


Assignment Problems Hungarian method

Step 1Balance the problem if it is unbalancedPlace an M as the cost element if some assignment is prohibitedConvert into equivalent min problem if it is a max problem

In a given matrix subtract the smallest element in each row from every element of that row and do the same in the column.


Step2 In the reduced matrix obtain from step 1, subtract the smallest element in each column from every element of that column

Step 3 Make the assignment for the reduced matrix obtained from step 1 and step 2

(all the zeros in rows/columns are either marked (□) or (x) and there is exactly one assignment in each row and each column. In such a case optimum assignment policy for the given problem is obtained.

If there is row or column with out an assignment go to the next step.


Step 4 Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure.

(i) mark(√) all rows that do not have assignments (ii) Mark (√) all columns (not already marked) which have

zeros in the marked rows (iii) Mark (√) all rows (not already marked) that have

assignments in marked columns (iv) Draw straight lines through all unmarked rows and

marked columns


Step 5 If the number of lines drawn are equal to the number of rows or columns, then it is an optimum

solution ,otherwise go to step 6 Step 6 Select the smallest element among all the

uncovered elements. Subtract this smallest element from all the uncovered elements an add it to the element which lies at the intersection of two line. Thus we obtain another reduced matrix for fresh assignments.

Step 7 go the step 3 and repeat the procedure until the umber of assignment become equal to the number of rows or columns. In such a case, we shall observe that row/column has an assignment. Thus, the current solution is an optimum solution.


Problem 1 A company centre has got four expert

programmers. The centre needs four application programmes to be developed. The head of the computer centre, after studying carefully the programme’s to be developed, estimate the computer time in minutes required by the respective experts to develop the application programmes as follows.


Programmers

A B C D

1 120 100 80 90

2 80 90 110 70

3 110 140 120 100

4 90 90 80 90


Problem 2 Suggest optimum solution to the following assignment

problem and also the minimum cost:


Markets/salesmen

I II III IV

A 44 80 52 60

B 60 56 40 72

C 36 60 48 48

D 52 76 36 40



Transportation

Transportation models deals with the transportation of a product manufactured at different plants or factories supply origins) to a number of different warehouses (demand destinations). The objective to satisfy the destination requirements within the plants capacity constraints at the minimum transportation cost.


A typical transportation problem contains

Inputs:Sources with availability

Destinations with requirementsUnit cost of transportation from various sources to destinations

Objective:To determine schedule of transportation to minimize total transportation cost


How to solve?1. Define the objective function to be

minimized with the constraints imposed on the problem.

2. Set up a transportation table with m rows representing the sources and n columns representing the destination

3. Develop an initial feasible solution to the problem by any of these methods a) The North west corner rule b) Lowest cost entry method c)Vogel’s approximation method


4. Examine whether the initial solution is feasible or not.( the solution is said to be feasible if the solution has allocations in ( m+n-1) cells with independent positions.

5. Test wither the solution obtained in the above step is optimum or not using a) Stepping stone method b) Modified distribution (MODI) method.

6.If the solution is not optimum ,modify the shipping schedule. Repeat the above until an optimum solution is obtained.


Applications

To minimize shipping costs from factories to warehouses or from warehouses to retails outlets.

To determine lowest cost location of a new factor, warehouse or sales office

To determine minimum cost production schedule that satisfies firm’s demand and production limitations.


North West corner method

1. Select the northwest corner cell of the transportation table and allocate as many units as possible equal to the minimum between availability supply and demand requirements i.e.(min (s1,d1)

2. Adjust the supply and demand numbers in the respective rows and columns allocation

3. A. If the supply for the first row is exhausted ,then move down to the first cell in the second row and first column and go to step 2.

4. If the demand for the first column is satisfied, then move horizontally to the next cell in the second column and first row and go to step 2


5. If for any cell, supply equals demand, then the next allocation can be made in cell either in the next row or column.

6. Continue the procedure until the total available quantity is fully allocated to the cells as required.

Advantages; it is simple and reliable. Easy to compute ,understand and interpret.

Disadvantages: This method does not take into considerations the shipping cost, consequently the initial solution obtained b this method require improvement.


Problem 1: Obtain initial solution in the following transportation problem by using Northwest corner rule method

Origins D1 D2 D3 D4 Supply/capacity/availabili

ty

O1 1 2 1 4 3030

O2 3 3 2 1 5050

O3 4 2 5 9 2020Demand/Requirements

2020 4040 3030 1010


Least cost method:

1. Select the cell with the lowest transportation cost among all the rows or column of the transportation table

2. If the minimum cost is not unique, then select arbitrarily any cell with this minimum cost. (preferably where maximum allocation is possible)

3. Repeat steps 1 and 2 for the reduced table until the entire supply at different factories is exhausted to satisfy the demand at different warehouses.


Problem 1: Obtain initial solution in the following transportation problem by using Least cost method


ty

O1 1 2 1 4 3030

O2 3 3 2 1 5050


2020 4040 3030 1010


Vegel’s Approximation Method (VAM)1. Compute a penalty for each row and column in the

transportation table. The penalty for a given row and column is merely the difference between the smallest cost and next smallest cost in that particular row or column.

2. Identify the row or column with the largest penalty. In this identified row or column, choose the cell which has the smallest cost and allocate the maximum possible quantity to the lowest cost cell in that row or column so as to exhaust either the supply at a particular source or satisfy demand at warehouse.( If a tie occurs in the penalties, select that row/column which has minimum cost. If there is a tie in the minimum cost also, select the row/column which will have maximum possible assignments)


3.Reduce the row supply or the column demanded by the assigned to the cell

4.If the row supply is now zero, eliminate the row, if the column demand is now zero, eliminate the column, if both the row, supply and the column demand are zero, eliminate both the row and column.

5. Recompute the row and column difference for the reduced transportation table, omitting rows or columns crossed out in the preceding step.

6. Repeat the above procedure until the entire supply at factories are exhausted to satisfy demand at different warehouses.


Problem 1: Obtain initial solution in the following transportation problem by using VAM


ty

O1 1 2 1 4 3030

O2 3 3 2 1 5050


2020 4040 3030 1010


The Modified Distribution Method1. Determine an initial basic feasible solution consisting

of m + n -1 allocations in independent positions using any of the three methods

2. Determine a set of number for each row and each column. Calculate Ui ( i= 1,2,..m)and Vj (j = 1,2..n)for

each column, and Cij = (Ui + Vj ) for occupied cells.

3. Compute the opportunity cost

Δij = Cij - (Ui + Vj ) for each unoccupied cells.


4. Check the sign of each opportunity cost: if all the

Δij are positive or zero, the given solution is optimum. If one of the values is zero there is another alterative solution for the same transportation cost. If any value is negative the given solution is not optimum. Further improvement is possible.

5. Select the unoccupied cell with the largest negative opportunity cost as the cell to be included in the next solution.

6. Draw a closed path or loop for the unoccupied cell selected in step 5.It may be noted that right angle turns in this path are permitted only a occupied cells and at the original unoccupied cell


7. Assign alternative plus and minus signs at the unoccupied cells on the corner points of the closed path with a plus sign at the cell being evaluated.

8.Determine the maximum number of units that should be shipped to this unoccupied cell. The smallest one with a negative position on the closed path indicates the number of units that can be shipped to the entering cell. This quantity is added to all the cells on the path marked with plus sign and subtract from those cells mark with minus sign. In this way the unoccupied cell under consideration becomes an occupied cell making one of the occupied cells as unoccupied cell.

9.Repeat the whole procedure until an optimum solution is

attained i.e. Δij is positive or zero. Finally calculate new transportation cost.


Problem: 4 A distribution system has the following constraints.

Factory capacity (in units) A 45 B 15 C 40 Warehouse Demand (in units) I 25 II 55 III 20 The transportation costs per unit( in rupees) allocated

with each route are as follows.


To/From

I II III

A 10 7 8

B 15 12 9

C 7 8 12


Find the optimum transportation schedule and the minimum total cost of transportation.

Problem 3 A company is spending Rs.1,000 on transportation of its

units from these plants to four distribution centres. The supply and demand of units, with unity cost of transporataion are given in the table.

What can be the maximum saving by optimum scheduling.


To/fro

m

D1 D2 D3 D4 Available

P1 19 30 50 12 7

P2 70 30 40 60 10

P3 40 10 60 20 18

Requiremetn

s

5 8 7 15


Special cases in Transportation Unbalanced transportation Maximisation Restricted routes


A product is produced by 4 factories F1, F2,F3 and F4. Their unit production cost are Rs.2,3,1,and 5 only. Production capacity of the factories are 50,70,40 and 50 units respectively. The product is supplied to 4 stores S1,S2,S3 and S4., the requirements of which are 25,35,105 and 20 respectively. Unit cost of transportation are given below


S1 S2 S3 S4

F1 2 4 6 11

F2 10 8 7 5

F3 13 3 9 12

F4 4 6 8 3


Find the optimal transportation plan such that total production and transportation cost is minimum.

PROBLEM A particular product is

manufactured in factories A,B ,C and D: it is sold at centres 1,2,and 3. the cost in rupees of product per unit and capacity of each plant is given below

Factory Cost Rs Per unit

Capacity

A 12 100

B 15 20

C 11 60

D 13 80


The sales prices is Rs per unit and the demand are as follows.

Find the optimal solution

Sales centers

Sales price per

unit

Demand

1 15 120

2 14 90

3 16 50


Assignment Problems Hungarian method

Step 1Balance the problem if it is unbalancedPlace an M as the cost element if some assignment is prohibitedConvert into equivalent min problem if it is a max problem

In a given matrix subtract the smallest element in each row from every element of that row and do the same in the column.


Step2 In the reduced matrix obtain from step 1, subtract the smallest element in each column from every element of that column

Step 3 Make the assignment for the reduced matrix obtained from step 1 and step 2

(all the zeros in rows/columns are either marked (□) or (x) and there is exactly one assignment in each row and each column. In such a case optimum assignment policy for the given problem is obtained.

If there is row or column with out an assignment go to the next step.


Step 4 Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure.

(i) mark(√) all rows that do not have assignments (ii) Mark (√) all columns (not already marked) which have

zeros in the marked rows (iii) Mark (√) all rows (not already marked) that have

assignments in marked columns (iv) Draw straight lines through all unmarked rows and

marked columns


Step 5 If the number of lines drawn are equal to the number of rows or columns, then it is an optimum

solution ,otherwise go to step 6 Step 6 Select the smallest element among all the

uncovered elements. Subtract this smallest element from all the uncovered elements an add it to the element which lies at the intersection of two line. Thus we obtain another reduced matrix for fresh assignments.

Step 7 go the step 3 and repeat the procedure until the umber of assignment become equal to the number of rows or columns. In such a case, we shall observe that row/column has an assignment. Thus, the current solution is an optimum solution.


Problem 1 A company centre has got four expert

programmers. The centre needs four application programmes to be developed. The head of the computer centre, after studying carefully the programme’s to be developed, estimate the computer time in minutes required by the respective experts to develop the application programmes as follows.


Programmers

A B C D

1 120 100 80 90

2 80 90 110 70

3 110 140 120 100

4 90 90 80 90


Problem 2Suggest optimum solution to the following assignment problem and also the minimum cost:


Markets/salesmen

I II III IV

A 44 80 52 60

B 60 56 40 72

C 36 60 48 48

D 52 76 36 40


Problem: 3 A company has taken the third floor of a multi-storied

building for rent with a view to locate one of their zonal offices. There are five main rooms in this to be assigned to five managers. Each room has it sown advantages and disadvantages. Some have windows, some are closer to the wash rooms or to the canteen or secretarial pool. The rooms are of all different sizes and shapes. Each of the five managers were asked to rank their room preference amongst the rooms 301,302,303,304 and 305. Their preferences were recorded in a table as indicated below


Most of the managers did not list al the five rooms since they were not satisfied with some of these rooms and they have left these from the list. Assuming that their preferences can be quantified by numbers, find out as to which manager should be assigned to which room so that their total preference ranking is a minimum.

M1 M2 M3 M4 M5

302 302 303 302 301

303 304 301 305 302

304 305 304 304 304

301 305

302

303


Problem 4 A company plans to assign 5 salesmen to 5 districts in which it operates. Estimates of sales revenue in thousands of rupees for each salesman in different districts are given in the following table. In your opinion, what should be the placement of the salesmen if the objective is to maximize the expected sales revenue?.


Expected sales data district wise

Salesmen

D1 D2 D3 D4 D5

S1 40 46 48 36 48

S2 48 32 36 29 44

S3 49 35 41 38 45

S4 30 46 49 44 44

S5 37 41 48 43 47


Problem5 : A traveling salesman has to visit 5 cities. He wishes to start from a particular city, visit each city once and then return to his starting point. The traveling cost for each city from a particular city is given below


What is the sequence of visit of the salesman, so that the cost is minimum.

From/to

A B C D E

A X 4 7 3 4

B 4 X 6 3 4

C 7 6 X 7 5

D 3 3 7 X 7

E 4 4 5 7 X


Problem6. A solicitors firm employs typists on hourly piece rate basis for their work . There are five typists for service and their charges and speeds are different. According to an earlier understanding only one job is given to one typist and the typist is paid for full hour even if he works for a fraction of an hour. Find the least cost allocation for the following data.


typist Rate per hour in Rs.

No.of pages typed per hour

A 5 12

B 6 14

C 3 8

D 4 10

E 4 11

JOB No. of pages

P 199

Q 175

R 145

S 198

T 178


Problem 7ABC airline, operating 7 days a week, has given the following time-table. The crews must have a minimum lay-over of 5 hours between flight. Obtain the pairing of flights that minimizes la-overtime away from home . For any given pairing the crew will be based at the city that results the smallest lay-over.


Hyderabad-Delhi Delhi -Hyderabad

Flight no.

Departure

Arrival Flight no. Departure

Arrival

A1 6 AM 8 AM B1 8 AM 10 AM

A2 8 AM 10 AM B2 9 AM 11AM

A3 2 PM 4 PM B3 2 PM 4 PM

A4 8 PM 10 PM B4 7 PM 9 PM


Problem 8. A company has four territories and four salesmen

available for assignment. The territories are not equally rich in their sales potential. It is estimated that a typical salesman operating I each territory would bring the following weekly sales.

Territory I II III IV

Annual sales

(in Rs)

60,000 50,000 40,000 30,000


The four salesmen are also considered to differ in ability. It is estimated that working under the same conditions their yearly sales would be proportionately as follows

Salesman A B C D

Proportion 7 5 5 4


If the criteria is to maximize expected sales, the intuitive answer is to assign the best salesman to the richest territory, the next best salesman to the second richest and so on. Verify this answer by the assignment technique.


Thank you Give feedback to

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Network Analysis

•Network is a graphical representation of all the

Activities and Events arranged in a logical and

sequential order.

•Network analysis plays an important role in project

management.

•A project is a combination of interrelated activities

all of which must be executed in a certain order for its

completion.


Activity: Activity is the actual performance of the

job. This consumes resources (Time, human

resources, money, and material)

Event: An event refers to start or completion of a

job. This does not consume any resources.


•Analyzing network, the planning, scheduling and

control of a project becomes easier.

•PERT and CPM are the two most popular network

analysis technique used to assist managers in

planning and controlling large scale projects.

•PERT- (Programme Evaluation Review Technique)

•CPM - (Critical Path Method)


Applications: -

Construction of a Residential complex,

Commercial complex,

Petro-chemical complex

Ship building

Satellite mission development

Installation of a pipe line project etc...


Historical Evolution.Before 1957 there was no generally accepted procedure to aid the management of a project. In 1958 PERT was developed by team of engineers working on a Polaris Missile programmer of the navy. This was a large project involved 250 prime contractors and about 9000 job contractors. It had about 19 million components. In such projects it is possible that a delay in the delivery of a small component might hold the progress of entire project. PERT was used successfully and the total time of completion was reduced from 7 years to 5 years.


In 1958 Du Pont Company used a technique called

Critical Path Method (CPM) to schedule and control a

very large project like overhauling of a chemical plant,

there by reducing the shutdown period from 130hrs to

90 hrs saving the company 1 million dollar.

Both of these techniques are referred to as project

scheduling techniques.


PERT CPM

1. It is a technique for planning scheduling & controlling of projects whose activities are subject to uncertainty in the performance time. Hence it is a probabilistic model

1. It is a technique for planning scheduling & controlling of projects whose activities not subjected to any uncertainty and the performance times are fixed. Hence it is a deterministic model

2.It is an Event oriented system

2. It is an Activity oriented system


3.Basically does not differentiate critical and non-critical activities

Differentiates clearly the critical activities from the other activities.

4. Used in projects where resources (men, materials, money) are always available when required.

4. Used in projects where overall costs is of primarily important. Therefore better utilized resources

5. Suitable for Research and Development projects where times cannot be predicted

5.Suitable for civil constructions, installation, ship building etc.


Rules for drawing the network diagrams.

•In a network diagram, arrows represent the

activities and circles represent the events.

•The tail of an arrow represents the start of an

activity and the head represent the completion

of the activity.

1 2 3 4


•The event numbered 1 denotes the start of the

project and is called initial event.

• Event carrying the highest number in the network

denotes the completion of the project and is called

terminal event.

1 2 3 4


•Each defined activity is represented by one and only

arrow in the network.

•Determine which operation must be completed

immediately before other can start.

•Determine which other operation must follow the

other given operation.

1 2 3 4


•The network should be developed on the basis of

logical, analytical and technical dependencies

between various activities of the project.

1 2 3 4


The basic network construction – Terminology used.

Network representation: There are two types of

systems –

AOA system (Activity on Arrow system)

AON system(Activity on Node system )

In this activities are represented by an arrows.

In this method activities are represented in the circles.


Problem 1.Construct an arrow diagram for the following project.

Activities Relationship

A Precedes B,C

B Precedes D,E

C Precedes D

D Precedes F

E Precedes G

F Precedes G


A

B

C

D

E

F

G



Job Immediate predecessor

Duration

A - 14 Days

B A 3 Days

C A 7 Days

D C 4 Days

E B,D 10 Days


A

B

C

D

E

14

3

7

4

10

KeyJob

Duration



Job Immediate predecessor

A -

B -

C A,B

D A

E D

F C,E


A

BC

D

E

F



Activity Predecessor

A -

B -

C -

D A,B

E B,C


A

B

C

D

E



Activity Predecessor

A -

B -

C -

D A,B

E B,C

F A,B,C


A

B

C

D

E

F


Problem 6.

Draw the PERT network for the following project

Event A is followed by events B & C

Event D is preceded by events B & C

Event H is the successor to event E

Event E is the successor to event B

Event F is the successor to event D & G

Event C is the predecessor to event G

Event J is preceded by events F,G, & H


A

B

E

D

C

G

H

F

J


Network Analysis



sequential order.


management.



completion.


Project situation (An example)

A new machine is required by a department for which budget approval is needed. The new machine necessitates employment of an operator who would be trained for operating this machine. The operator can be hired as soon as the proposal of buying machine is cleared, and trained on the same machine in the training division of the company. Once the machine is installed and worker is trained , the trail production can commence.


Activity Description Duration (weeks) Immediate Predecessor(s)

A Obtain the budget approval

2 -

B Obtain the machine 5 A

C Hire the operator 1 A

D Install the machine 1 B

E Train the operator 6 C

F Produce the goods 1 D,E

Problem Presentation


1 2 5 6

3

4

A

B

C

D

E

F

Obtain the machine

Obtain the budget approval

Hire the operator

Install the machine

Train the operator

Produce the goods


1 2A



1 2

3

A

B

Obtain the machine




1 2

3

4

A

B

C

Obtain the machine


Hire the operator


1 2

3

4

A

B

C

D

Obtain the machine


Hire the operator

Install the machine

5


1 2 5

3

4

A

B

C

D

E

Obtain the machine


Hire the operator

Install the machine

Train the operator


1 2 5 6

3

4

A

B

C

D

E

F

Obtain the machine


Hire the operator

Install the machine

Train the operator

Produce the goods


Network Analysis



sequential order.


management.



completion.


•The network should be developed on the

basis of logical, analytical and technical

dependencies between various activities

of the project.


CRITICAL PATH

Meaning: The longest path in a project network which determine the duration of the project is known as critical path.


1

3

4

6

2

5

7

8

Determination of Critical Path

2

4

3

5

4

5

23

26

1

Step 1.List all the possible sequences from start to finish

Step 2.For each sequence determine the total time required from start to finish.

Step 3.Identify the longest path (Critical Path)


Step 1. List all the possible sequences from start to finish

Path A : 1 – 2 – 5 – 8

Path B : 1 – 3 – 5 – 8

Path C : 1 – 3 – 6 – 7 – 8

Path D : 1 – 3 – 4 – 7 – 8

Path E : 1 – 3 – 4 – 6 – 7 – 8

Step 2.For each sequence determine the total time required from start to finish.Path A : 2 + 3 + 4 = 9 daysPath B : 4 + 5 + 4 = 13 daysPath C : 4 + 5 + 6 + 1 = 16 daysPath D : 4 + 2 + 3 + 1 = 10 daysPath E : 4 + 2 + 2 + 6 + 1 = 10 days


Step 2.For each sequence determine the total time

required from start to finish.

Path A : 2 + 3 + 4 = 9 days

Path B : 4 + 5 + 4 = 13 days

Path C : 4 + 5 + 6 + 1 = 16 days

Path D : 4 + 2 + 3 + 1 = 10 days

Path E : 4 + 2 + 2 + 6 + 1 = 10 days


Path C : 4 + 5 + 6 + 1 = 16 days

Path C : 1 – 3 – 6 – 7 – 8


1

3

4

6

2

5

7

8

Determination of Critical Path

2

4

3

5

4

5

23

26

1

Step 1.List all the possible sequences from start to finish

Step 2.For each sequence determine the total time required from start to finish.



Float (Slack)

•Float (Slack ) refers to the amount of time by which a particular event or an activity can be delayed without affecting the time schedule of the network.

•Float (Slack)

Float (Slack) is defined as the difference between latest allowable and the earliest expected time.

Event Float/Slack = LS – ES

Where LS = Latest start time

ES = Early start time.


Earliest start : Denoted as ‘ES’

Earliest start time is the earliest possible time by

which the activity can be started.

Early finish time : Denoted as ‘EF’

Early finish time is the earliest possible time by

which the activity can be completed.


Latest start time : Denoted as ‘LS’

Latest start time is the latest possible time by

which the activity can be started

Late finish time : Denoted as ‘LS’

Late finish time is the latest possible time by which

the activity can be completed.


Total float (TF) / Total slack (TS)

Total float of the job is the differences between its Late start and Early start ‘or’ Late finish and Early finish

i.e.

TF( CA) = LS (CA) - ES (CA)

Or

TF( CA) = LF (CA) - EF (CA)

CA = Current activity


Free float (FF)

Free float is the amount of time a job can be delayed without affecting the Early start time of any other job.

FF(CA) = ES(SA) – EF (CA)

CA = Current Activity

SA = Succeeding Activity


Independent Float (IF)

Independent Float is the amount of time that can be delayed without affecting either predecessor or successor activities.

IF = ES(SA) – LF(PA) - Duration of CA

ES = Early Start

LF = Late Finish


PA = Preceding Activity



Example:

Construct the Network for the following

Project and determine the following

i) Critical Path

ii) ES,EF,LS,LF

iii) TF,FF


Activity Duration

1-2 14

1-4 3

2-3 7

2-4 0

3-5 4

4-5 3

5-6 10


1

4

6

3

5

2

B

3

C

3

A14

D

7

E4

F

10G 0

Construction of the Network and Determination Critical Path


1

4

6

3

5

2

B

3

C

3

A(0,14)14

D

7

E4

F

10G 0

Determination of ES and EF


1

4

6

3

5

2

B

3

C

3

A(0,14)14

D(14,21)

7

E4

F

10G 0



1

4

6

3

5

2

B

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

100G

F



1

4

6

3

5

2

B

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

10G 0



1

4

6

3

5

2

B(0,3)

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

100G



1

4

6

3

5

2

B(0,3)

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

10G

(14,

14)

0



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

10G

(14,

14)

0

F(25,35)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

10(25,35)G

(14,

14)

Determination of LS and LF


1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3(22,25)

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3(22,25)

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)



1

4

6

3

5

2

B(0,3)

3(19,22)

C(14,17)3(22,25)

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)

KeyJob (ES,EF)

Duration (LS,LF)



1

4

6

3

5

2

B(0,3)

3(19,22)

C(14,17)3(22,25)

A(0,14)(0)14 (0,14)(0) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)

Determination of TF and FF

G(1

4,14

)

0(22

,22)



IF = ( ES(SA) – LF(PA)) - Duration of CA

KeyJob (ES,EF)(FF)

Duration (LS,LF)(TS)


1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)3(22,25)

A(0,14)(0)14 (0,14)(0) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)(0)

7(14,21)(0)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)(0)

7(14,21)(0)

E(2

1,25

)(0)

4(21

,25)

(0)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)(0)

7(14,21)(0)

E(2

1,25

)(0)

4(21

,25)

(0)

F(25,35)(0)

10(25,35)(0)0(

22,2

2)

G(1

4,14

)





KeyJob (ES,EF)(FF)



Activity Duration ES EF LS LF TF FF

1-2 14 0 14 0 14 0 0

1-4 3 0 3 19 22 19 11

2-3 7 14 21 14 21 0 0

2-4 0 14 14 22 22 0 0

3-5 4 21 25 21 25 0 0

4-5 3 14 17 22 25 8 8

5-6 10 25 35 25 35 0 0


Example:



i) Critical Path

ii) ES,EF,LS,LF

iii) TF,FF


Activity Duration

1-2 2

2-3 3

2-4 5

3-5 4

3-6 1

4-6 6

4-7 2

5-8 8

6-8 7

7-8 4


1 2

4 7

3

6

5

8

(0, 2)(0)

(ES,EF)(FF)

Duration (LS,LF)(TS)Key

2(0, 2)(0)

(2, 5)(0)

3(5, 8)(3)

(5 ,9)(0

)

4(8, 12)(3)

(2, 7)(0)5(2, 7)(0)

(5, 6)(7)1(12, 13)(7)

(7,13) (0)

6(7, 13)(0

)

(7, 9)(0)

2(14,16)(7)

(13, 20)(0)

7(13, 20)(0)

(9,1

3)(0

)4(

16, 2

0)(7

)

(9,17)(0)8(12,20)(3)



1-2 2 0 2 0 2 0 0

2-3 3 2 5 2 8 3 0

2-4 5 2 7 5 7 0 0

3-5 4 5 9 8 12 3 0

3-6 1 5 6 12 13 7 7

4-6 6 7 13 7 13 0 0

4-7 2 7 9 14 16 7 0

5-8 8 9 17 12 20 3 0

6-8 7 13 20 13 20 0 0

7-8 4 9 13 16 20 7 0


PERT Model

Historical Evolution.Before 1957 there was no generally accepted procedure to aid the management of a project. In 1958 PERT was developed by team of engineers working on a Polaris Missile programme of the navy. This was a large project involved 250 prime contractors and about 9000 job contractors. It had about 19 million components. In such projects it is possible that a delay in the delivery of a small component might hold the progress of entire project. PERT was used successfully and the total time of completion was reduced from 7 years to 5 years.


Differences between PERT & CPM

PERT CPM














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Transportation

Transportation models deals with the transportation of a product manufactured at different plants or factories supply origins) to a number of different warehouses (demand destinations). The objective to satisfy the destination requirements within the plants capacity constraints at the minimum transportation cost.


A typical transportation problem contains

Inputs:Sources with availability

Destinations with requirementsUnit cost of transportation from various sources to destinations

Objective:To determine schedule of transportation to minimize total transportation cost


How to solve?

1. Define the objective function to be minimized with the constraints imposed on the problem.

2. Set up a transportation table with m rows representing the sources and n columns representing the destination

3. Develop an initial feasible solution to the problem by any of these methods a) The North west corner rule b) Lowest cost entry method c)Vogel’s approximation method


4. Examine whether the initial solution is feasible or not.( the solution is said to be feasible if the solution has allocations in ( m+n-1) cells with independent positions.

5. Test wither the solution obtained in the above step is optimum or not using a) Stepping stone method b) Modified distribution (MODI) method.

6.If the solution is not optimum ,modify the shipping schedule. Repeat the above until an optimum solution is obtained.


Applications

To minimize shipping costs from factories to warehouses or from warehouses to retails outlets.

To determine lowest cot location of a new factor, warehouse or sales office

To determine minimum cost production schedule that satisfies firm’s demand and production limitations.


North West corner method1. Select the northwest corner cell of the transportation

table and allocate as many units as possible equal to the minimum between availability supply and demand requirements i.e.(min (s1,d1)

2. Adjust the supply and demand numbers in the respective rows and columns allocation

3. A. If the supply for the first row is exhausted ,then move down to the first cell in the second row and first column and go to step 2.

4. If the demand for the first column is satisfied, then move horizontally to the next cell in the second column and first row and go to step 2


5. If for any cell, supply equals demand, then the next allocation can be made in cell either in the next row or column.

6. Continue the procedure until the total available quantity is fully allocated to the ells as required.

Advantages; it is simple and reliable. Easy to compute ,understand and interpret.

Disadvantages: This method does not take into considerations the shipping cost, consequently the initial solution obtained b this method require improvement.




ty

O1 1 2 1 4 3030

O2 3 3 2 1 5050


2020 4040 3030 1010


Least cost method:

1. Select the cell with the lowest transportation cost among all the rows or column of the transportation table

2. If the minimum cost is not unique, then select arbitrarily any cell with this minimum cost. (preferably where maximum allocation is possible)

3. Repeat steps 1 and 2 for the reduced table until the entire supply at different factories is exhausted to satisfy the demand at different warehouses.




ty

O1 1 2 1 4 3030

O2 3 3 2 1 5050


2020 4040 3030 1010


Vegel’s Approximation Method (VAM)1. Compute a penalty for each row and column in the

transportation table. The penalty for a given row and column is merely the difference between the smallest cost and next smallest cost in that particular row or column.

2. Identify the row or column with the largest penalty. In this identified row or column, choose the cell which has the smallest cost and allocate the maximum possible quantity to the lowest cost cell in that row or column so as to exhaust either the supply at a particular source or satisfy demand at warehouse.( If a tie occurs in the penalties, select that row/column which has minimum cost. If there is a tie in the minimum cost also, select the row/column which will have maximum possible assignments)


3.Reduce the row supply or the column demanded by the assigned to the cell

4.If the row supply is now zero, eliminate the row, if the column demand is now zero, eliminate the column, if both the row, supply and the column demand are zero, eliminate both the row and column.

5. Recompute the row and column difference for the reduced transportation table, omitting rows or columns crossed out in the preceding step.

6. Repeat the above procedure until the entire supply at factories are exhausted to satisfy demand at different warehouses.


Theory of games

Game theory may be defined as “ a body of knowledge that deals with making decisions when two or more intelligent and rational opponents are involved under conditions of conflict and competition.

The approach of the game theory is to seek to determine a rival’s most profitable counter-strategy to one’s own ‘best moves to formulate the appropriate defensive measures.


Basic terminology Game: Let us consider a two person coin tossing

game between two player X and Y . Each player tosses an unbiased coin simultaneously. If the two simultaneous tosses match i.e. the toss of each player given either heard up or the tail up simultaneous ( eg. Either (H,H ) or (T,T) the player Y pays Rs.100 to player X. But if the toss of player x gives Head up and toss of the player Y gives tail up or vice versa, then the player X pays Rs.80 to Y. This can be put up in the tabular form as follows


Y

X

Y pays Rs.100 to X

X pays Rs.80 t0 Y

X pays Rs.80 t0 Y

Y pays Rs.100 to X


In this game of coin tossing, the outcome of the toss of each player is only two i.e. Hear or Tail, which are called the moves or strategies of the players. Here the strategies of both the players are the same but they may be different. The results or termination of each play., is expressed in terms of X’ s payoff with the help of payoff matrix as shown in the above table.


A competitive situation is thus called a ‘game’ if it has the following properties.

The number of competitors called players is finite

The players at rationally and intelligently Each player has available to him a finite

number of choices or possible course of actions called strategies. The number of choices need not be the same for each player.


All relevant information, i.e., the different strategies of each player and the amount of gain or loss on an individual's move( strategy) are known to each player in advance.

The players select their respective courses of action ( strategies) simultaneously.

The player make individual decisions without direct communication

The maximizing player attempt to maximize his gains and the minimizing player tries to minimize his losses

The pay-off is fixed and determined in advance.


Two persons zero sum game: a game of two persons, in which the gains of one player are the losses of the other player is called a two person zero sum game, i.e., in a two person zero sum game, the algebraic sum of the gains to both the players after a play is bound to be zero. These are also known as rectangular games.


Pay off matrix: A strategy is a course of action taken by one of the participants in a game, and the payoff is the result or outcome of the strategy.

Maximin Principle: A player adopts a pessimistic attitude and plays safe. i.e. his strategy is always that which result in the best out of the worst outcomes. HE decides to play that strategy which corresponds to the maximum of the minimum gains for different course of actions.


Minimax Principle: The minimizing player would also like to play safe and he selects the strategy which corresponds to the minimum of the maximum losses for his different course of action.

Optimal strategy: A course of action or play which puts the player in the most preferred position, irrespecctive of the strategy of his competitors is called an optimal strategy.


Value of the game: It is the expected pay-off of play when all the players of the game follow their optimal strategies. The game is called fair if the value of the game is zero and unfair if it is non-zero.

Solutions methods of pure strategy games (with saddle point)

in case of pure strategy game, the maximizing player arrives at his optimal strategy o the basis of the miximin criterion, while the minimizing player’s strategy is based on the minimax criterion. The game is solved when the maximin value equals minimax value( which is known as saddle point)


1. Develop the payoff matrix2. Identify row minimums and select the

largest of these as player one’s maximin strategy

3. Identify column maximums and select the smallest of these as the opponent’s minimax strategy.

4. If the maximin value equals the minimax value, the game is a pure strategy game and that value is the saddle point.

5. The value of the game of player one is the maximin value, and to player two, the value is the negative of the minimax value.


Problem 1: Solve the following games whose payoff matrix is given by

Player B

Player

A

B1 B2 B3 B4

A1 1 7 3 4

A2 5 6 4 5

A3 7 2 0 3


Firm B

Firm A

B1 B2 B3 B4 B5

A1 3 -1 4 6 7

A2 -1 8 2 4 12

A3 16 8 6 14 12

A4 1 11 -4 2 1


PRINCIPLE OF DOMINANCE

If a strategy is inferior to another, it is said to be dominated

1. If all elements in a row are less than or equal to the corresponding elements in another row, then that row is dominated and can be deleted from the matrix.

2. If all elements in a column are grater than or equal to the corresponding elements in another column, then that column is dominated and can be deleted from the matrix.

3. A pure strategy may be dominated if if is inferior to average of two or more other pure strategies.


Note: dominance property is used when the saddle point does not exist

Problem. Apply the rule of dominance for the following matrix .

Player B

Player

A

B1 B2 B3

A1 9 8 -7

A2 3 -6 4

A3 6 7 7


Solution methods of strategy games (games without saddle point)


If it is a 2 × 2 game Player B b1 b2

Player a1 a11 a12 A a2 a21 a22

If A plays a1 with probability x and a2 with probability 1-x, and B plays b1 with probability y and b2 with probability 1-y, then

a22 – a21 a22 – a12

x = ----------------------------- y = --------------------------- (a11 + a22) – (a21 + a12) (a11 + a22) – (a21 + a12)and (a11 × a22) – (a21 × a12) v = ----------------------------- (a11 + a22) – (a21 + a12)


Pure strategy: If a player knows exactly what the other player is going to do, a deterministic situation is obtained and objective function is to maximize the gain. Therefore, the pure strategy is a decision rule always to select a particular course of action.

Mixed strategy: If a player is guessing as to which activity is to be selected by the other on any particular occasion, a probabilistic situation is obtained and objective function is to maximize the expected gain. Thus, the mixed strategy is a selection among pure strategies with fixed probabiities.


Problem 4 Solve the following game

Player B

Player

A

I II III IV

I -5 2 1 20

II 5 5 4 6

III 4 -2 0 -5


Problem 5:Two players A and B without showing each other, put on a table a coin, with head or tail up. A wins Rs.8 when both the coins show head and Rs.1 When both are tails. B wins Rs.3 when the coins do not match. Given the choice of being matching player ( A) or non-matching player (B) which one would you choose and what would be your strategy?


Problem 6

Company B

company A

I II III IV

I 3 2 4 0

II 3 4 2 4

III 4 2 4 0

IV 0 4 0 8


Problem 7 Two firms A and B make colour and black& white T.V. sets. Firm A can make either 150 colour sets in a week or an equal number of B & W T.V. sets, and make a profit of Rs.400 per colour set and Rs.300 for B & W set. Firm B can, on the other hand make either 300 colour sets, or 150 colour and 150 B & W sets, or 300 B & W sets and manufacturers would share market in the proportion in which they manufacture a particular type of set. Write the pay off matrix of A per week. Obtain A’s and B’s optimum strategies and value of the game.


Problem 8 Graphical method Solve the following (2X 3) game graphically

Player B

Player

A

b1 b2 b3

a1 1 3 11

a2 8 5 2


Problem 9 Solve the following game graphically whose e

payoff matrix for the player A is given in the table

Player B

I II

Player A I 2 4

II 2 3

III 3 2

IV -2 6


Revision: Transportation

Given the following transportation problem

MARKET

WHAREHOUSE

A B C SUPPLY

W1 10 12 7 180

W2 14 11 6 100

W3 9 5 13 160

W4 11 7 9 120

Demand 240 200 220


Obtain the initial basic feasible solution by VAM. Is the optimal solution obtained by you unique? If not, what is the other optimal solution?


Assignment: Assign the mechanics to the jobs .

JOB

M

E

C

H

A

NIC

1 2 3 4 5

A 10 3 3 2 8

B 9 7 8 2 7

C 7 5 6 2 4

D 9 10 9 6 10


Model: 1Operating Characteristicsa) Queue length

average number of customers in queue waiting to get service

b) System length average number of customers in the system

c) Waiting time in queue average waiting time of a customer to get service

d) Total time in system average time a customer spends in the system

e) Server idle time relative frequency with which system is idle


Measurement parameters λ= mean number of arrivals per time period (eg.

Per hour) μ = mean number of customers served per time period Probability of system being busy/traffic intensity ρ = λ / μ Average waiting time system Ws = 1/(μ- λ) Average waiting time in queue Wq= λ/ μ(μ- λ) Average number of customers in the system Ls = λ/ (μ- λ)


Average number of customers in the queue Lq = λ2/ μ(μ- λ) Probability of an empty facility/system being idle P(0) = 1– P(w) Probability of being in the system longer than

time (t) P(T>t)= e –(μ- λ)t

Probability of customers not exceeding k in the system

P (n.≥k) = ρk

P( n>k) = ρ(k+1)

Probability of exactly N customers in the system P(N) = ρN (1-ρ)


Problem : At a service counter of fast-food joint, the customers arrive at the average interval of six minutes whereas the counter clerk takes on an average 5 minutes for preparation of bill and delivery of the item. Calculate the following

a. counter utilisation levelb. average waiting time of the customers at

the fast food jointc. Expected average waiting time in the line


d. Average number of customers in the service counter area

e. average number of customer in the linef. probability that the counter clerk is idleg. Probability of finding the clerk busyh. chances that customer is required to wait

more than 30 minutes in the systemi. probability of having four customer in the

systemJ) probability of finding more than 3 customer in

the system


Solutions: Given λ = 60/10 = 10 customer/hr

μ = 12 customer/hr A) traffic intensity ρ = λ / μ = 10/12 = 0.833 b) waiting time in the system Ws = 1/ μ- λ = 1/12-10 = 0.5 hr C) waiting time in the queue Wq = λ/ μ (μ- λ) = 10/12(12-10) = 0.416 hr D) number of customer in the system Ls= λ/ (μ- λ) = 10/12-10 = 5 customers E) Number of customer in the queue Lq = λ2 / μ (μ- λ) = 102 /12(12-10) = 4.167 customers


f) probability that the counter clerk is idle 1- ρ = 1- λ / μ = 1- 10/12 = 0.167 g. Probability of finding the clerk busy ρ = λ / μ = 10/12 = 0.833h) chances of probability that customer

wait more than 30min = 30/60 = 0.5 hrs

P (T>t) = e – (μ- λ) t

P (T>0.5) = e – (12- 10) 0.5 = 0.368


I) probability of having four customer in the system

P (N) = ρN (1-ρ) P (4) = ρ4 (1-ρ) = (0.833)4(1-0.833) = 0.0806

j) probability of finding more than 3 customer in the system

P (n>k) = ρ (k+1)

P (n>3) = ρ (3+1) = (λ / μ) 4= (10/12) 4 = 0.474


Queuing Theory

Queuing System: General Structure

Arrival Process According to source According to numbers According to time

Service System Single server facility Multiple, parallel facilities with single queue Multiple, parallel facilities with multiple queues Service facilities in a parallel


Queue Structure First come first served Last come first served Service in random order Priority service


Model 1: Poisson-exponential single server model – infinite population

Assumptions: Arrivals are Poisson with a mean arrival rate of, say

λ Service time is exponential, rate being μ Source population is infinite Customer service on first come first served basis Single service stationFor the system to be workable, λ ≤ μ


Model 2: Poisson-exponential single server model – finite populationHas same assumptions as model 1, except

that population is finite


Model 3: Poisson-exponential multiple server model – infinite population

Assumptions Arrival of customers follows Poisson law, mean rate λ Service time has exponential distribution, mean

service rate μ There are K service stations A single waiting line is formed Source population is infinite Service on a first-come-first-served basis Arrival rate is smaller than combined service rate of all

service facilities


Model: 1Operating Characteristicsa) Queue length

average number of customers in queue waiting to get service

b) System length average number of customers in the system

c) Waiting time in queue average waiting time of a customer to get service

d) Total time in system average time a customer spends in the system

e) Server idle time relative frequency with which system is idle


Measurement parameters λ= mean number of arrivals per time period (eg.

Per hour) μ = mean number of customers served per time period Probability of system being busy/traffic intensity ρ = λ / μ Average waiting time system Ws = 1/(μ- λ) Average waiting time in queue Wq= λ/ μ(μ- λ) Average number of customers in the system Ls = λ/ (μ- λ)


Average number of customers in the queue Lq = λ2/ μ(μ- λ) Probability of an empty facility/system being idle P(0) = 1– P(w) Probability of being in the system longer than time (t) P(T>t)= e –(μ- λ)t

Probability of customers not exceeding k in the system P (n.≥k) = ρk

P( n>k) = ρ(k+1)

Probability of exactly N customers in the system P(N) = ρN (1-ρ)


Problem 1. Customers arrive at a booking office window, being manned by a single individual a a rate of 25per hour. Time required to serve a customer has exponential distribution with a mean of 120 seconds. Find the mean waiting time of a customer in the queue.


Problem 2: A repairman is to be hired to repair machines which breakdown at a n average rate of 6 per hour. The breakdowns follow Poisson distribution. The non-production time of a machine is considered to cost Rs. 20 per hour. Two repairmen Mr. X and Mr.Y have been interviewed for this purpose. Mr. X charges Rs.10 per hour and he service breakdown machines at the rate of 8 per hour. Mr. Y demands Rs.14 per hour and he services at an average of 12 per hour. Which repairman should be hired? ( Assume 8 hours shift per day)


Problem 3: A warehouse has only one loading dock manned by a three person crew. Trucks arrive at the loading dock at an average rate of 4 trucks per hour and the arrival rate is Poisson distributed. The loading of a truck takes 10 minutes on an average and can be assumed to be exponentially distributed . The operating cost of a truck is Rs.20 per hour and the members of the crew are paid @ Rs.6 each per hour. Would you advise the truck owner to add another crew of three persons?


Problem 4; At a service counter of fast-food joint, the customers arrive at the average interval of six minutes whereas the counter clerk takes on an average 5 minutes for preparation of bill and delivery of the item. Calculate the following

a. counter utilisation levelb. average waiting time of th4e customers

at the fast food jointc. Expected average waiting time in the line


d. Average number of customers in the service counter area

e. average number of customer in the linef. probability that the counter clerk is idleg. Probability of finding the clerk busyh. chances that customer is required to wait

more than 30 minutes in the systemi. probability of having four customer in the

systemJ) probability of finding more than 3 customer in

the system


Problem 5: Customers arrive at a one-window drive-in bank according to a Poisson distribution with mean 10 per hour. Service time per customer is exponential with mean 5 minutes. The space in front of the window including that for the serviced car accommodate a maximum of 3 cars. Other cars can wait outside the space. Calculate

A) what is the probability that an arriving customer can drive directly to the space in front of the window.

B) what is the probability that an arriving customer will have to wait outside the indicated space

C) How long is arriving customer expected to wait before stating the service.


D) How many spaces should be provided in front of the window so that all the arriving customers can wait in front of the window at least 20% of the time.

Problem 6Customers arrive at the first class ticket counter of a

theatre at a rate of 12 per hours. There is one clerk serving the customers at a rate of 30 per hour. Assuming the conditions for use of the single channel queuing model, evaluate


a) The probability that there is no customer at the counter (i.e. that the system is idle)

b) The probability that there are more than 20 customers at the counter

c) The probability that there is no customer waiting to be served

d) The probability that a customer is being served and no body is waiting.



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Replacement decisions

Replacement theory is concerned with the problem of replacement of machines, electricity bulbs etc due to their deteriorating efficiency ,failure or breakdown. Replacement is generally carried out in the following situation. When the existing items outlived, when existing items destroyed


Failure mechanism of items A) Gradual failure( increase in expenditure for

operating costs, decreased in productivity of the equipment

B) Sudden failure i) Progressive failure ii) Retrogressive failure iii) Random failure


Problem1: The cost of a machine is Rs 6100 & its scrap value is only Rs 100. Maintenance costs are followed from the experience. When should the machine be replaced

Year 1 2 3 4 5 6 7 8

Maintenace cost

100 250 400 600 900 1250 1600 2000


Problem 2

Year

1 2 3 4 5 6 7 8

Maintenace cost

1000

1200 1400 1800 2300 2800 3400 4000

Resale price

3000 1500 750 375 200 200 200 200


Replacement of machines that deteriorates with time (considering time value of money)

1. Write in a column the running/maintenance costs of machine or equipment for different years Rn

2. In the next column write the discount factor indicating the present value of a rupee received after (i-1)year., v n-1

=(1/1+r ) n-1


3.The two column values are multiplied to get present value of the maintenance costs i.e. Rn v n-1

4.These discounted maintenance costs are then cumulated to the ith year to get

∑Rn v n-1

5.The cost of the machine or equipment is added to the values obtained in step 4 above to obtain C + ∑Rn v n-1

6. The discount factors are then cumulated to get ∑ v n-1


7. The total costs obtained in (step 5) are divided by the corresponding value of the accumulated discount factor for each of the years

8. Now compare the column of maintenance costs which is constantly increasing, with the last column. Replace the machine in the latest year that the last column exceeds the column of maintenance cost.


Problem. The cost of a new machine is Rs.5000. The maintenance cost during the nth year is given by Rn = 500 X (n-1), n=1,2,…Suppose that the discount rate per year is 0.05 . After how many years it will be economical to replace the machine by a new one?


Problem An entrepreneur is considering to purchase a

machine for his factory. Relevant data about alternative machines are as follows.

As a advisor to the buyer, you have been asked to select the best machine considering 12%normal rate of return.

Single payment present worth factor at 12%for 10 years 0.322

Annual series present worth factor at 12% for 10 years 5.650


machine A B C

Present investment.

10,000 12000 15000

Total annual cost

2000 1500 1200

life 10 10 10

Salvage value

500 1000 1200


Problem 4: The management of a large hotel is considering the periodic replacement of light bulbs fitted in ins rooms. There are 500 rooms in the hotel and each room has 6 bulbs. The management is now following the policy of replacing the bulbs as they fail at a total cost of Rs.3 per bulb. The management feels that this cost can be reduced to Rs.1 by adopting the periodic replacement method. On the basis of the information give below, evaluate the alternative and make a recommendation to the management.


Months of use 1 2 3 4 5

% of bulbs failing by that month

10 25 50 80 100


Assumptions 1. Bulbs failing during a month are replaced just before the end of the month.

The actual percentage of failures during a month as for a sub population of bulbs with same age is the same as the expected percentages of failures during the month for that sub population


Problem 5:Following mortality rates have been observed for a certain type of fuses

Week 1 2 3 4 5

%of failure at the end of week

5 15 35 75 100


There are 1,000 fuses in use and it costs Rs.5 to replace an individual fuse. If all fuses were replaced simultaneously if would cost Rs.1.25 per fuse. It is proposed to replace all fuses at fixed intervals of time, whether or not they have burnt out, and to continue replacing burnt-out fuses as they fail. At what intervals, the group replacement should be made? Also prove that this optimal policy is superior to the straight forward policy of replacing each fuse only when it fails.


Problem 6: A computer contains 10,000 resistors. When any of the register fails it is replaced. The cost of replacing a single resistor is Rs.10 only. If all the resistors are replaced at the same time, the cost per resistor would be reduced to Rs.3.50. The per cent surviving by the end of the month t is as follows. What is the optimum plan.


month 0 1 2 3 4 5 6

%surviving at the end of month

100 97 90 70 30 15 0


Float (Slack)

•Float (Slack ) refers to the amount of time by which a particular event or an activity can be delayed without affecting the time schedule of the network.

•Float (Slack)

Float (Slack) is defined as the difference between latest allowable and the earliest expected time.

Event Float/Slack = LS – ES

Where LS = Latest start time

ES = Early start time.


Earliest start : Denoted as ‘ES’

Earliest start time is the earliest possible time by

which the activity can be started.

Early finish time : Denoted as ‘EF’

Early finish time is the earliest possible time by

which the activity can be completed.


Latest start time : Denoted as ‘LS’

Latest start time is the latest possible time by

which the activity can be started

Late finish time : Denoted as ‘LS’

Late finish time is the latest possible time by which

the activity can be completed.


Total float (TF) / Total slack (TS)

Total float of the job is the differences between its Late start and Early start ‘or’ Late finish and Early finish

i.e.


Or

TF( CA) = LF (CA) - EF (CA)

CA = Current activity


Free float (FF)

Free float is the amount of time a job can be delayed without affecting the Early start time of any other job.





Independent Float (IF)

Independent Float is the amount of time that can be delayed without affecting either predecessor or successor activities.

IF = ES(SA) – LF(PA) - Duration of CA

ES = Early Start

LF = Late Finish


PA = Preceding Activity



Example:



i) Critical Path

ii) ES,EF,LS,LF

iii) TF,FF


Activity Duration

1-2 14

1-4 3

2-3 7

2-4 0

3-5 4

4-5 3

5-6 10


1

4

6

3

5

2

B

3

C

3

A14

D

7

E4

F

10G 0

Construction of the Network and Determination Critical Path


1

4

6

3

5

2

B

3

C

3

A(0,14)14

D

7

E4

F

10G 0



1

4

6

3

5

2

B

3

C

3

A(0,14)14

D(14,21)

7

E4

F

10G 0



1

4

6

3

5

2

B

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

100G

F



1

4

6

3

5

2

B

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

10G 0



1

4

6

3

5

2

B(0,3)

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

100G



1

4

6

3

5

2

B(0,3)

3

C

3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

10G

(14,

14)

0



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

10G

(14,

14)

0

F(25,35)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7

E(2

1,25

)4

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14

D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3(22,25)

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)



1

4

6

3

5

2

B(0,3)

3

C(14,17)3(22,25)

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)



1

4

6

3

5

2

B(0,3)

3(19,22)

C(14,17)3(22,25)

A(0,14)14 (0,14) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)

KeyJob (ES,EF)

Duration (LS,LF)



1

4

6

3

5

2

B(0,3)

3(19,22)

C(14,17)3(22,25)

A(0,14)(0)14 (0,14)(0) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)


G(1

4,14

)

0(22

,22)




KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)3(22,25)

A(0,14)(0)14 (0,14)(0) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)

7(14,21)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)(0)

7(14,21)(0)

E(2

1,25

)4(

21,2

5)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)(0)

7(14,21)(0)

E(2

1,25

)(0)

4(21

,25)

(0)

F(25,35)

10(25,35)G

(14,

14)

0(22

,22)





KeyJob (ES,EF)(FF)



1

4

6

3

5

2

B(0,3)(11)

3(19,22)(1

9)

C(14,17)(8)3(22,25)(8)

A(0,14)(0)14 (0,14)(0) D(14,21)(0)

7(14,21)(0)

E(2

1,25

)(0)

4(21

,25)

(0)

F(25,35)(0)

10(25,35)(0)0(

22,2

2)

G(1

4,14

)





KeyJob (ES,EF)(FF)




1-2 14 0 14 0 14 0 0

1-4 3 0 3 19 22 19 11

2-3 7 14 21 14 21 0 0

2-4 0 14 14 22 22 0 0

3-5 4 21 25 21 25 0 0

4-5 3 14 17 22 25 8 8

5-6 10 25 35 25 35 0 0


Example:



i) Critical Path

ii) ES,EF,LS,LF

iii) TF,FF


Activity Duration

1-2 2

2-3 3

2-4 5

3-5 4

3-6 1

4-6 6

4-7 2

5-8 8

6-8 7

7-8 4


1 2

4 7

3

6

5

8

(0, 2)(0)

(ES,EF)(FF)

Duration (LS,LF)(TS)Key

2(0, 2)(0)

(2, 5)(0)

3(5, 8)(3)

(5 ,9)(0

)

4(8, 12)(3)

(2, 7)(0)5(2, 7)(0)

(5, 6)(7)1(12, 13)(7)

(7,13) (0)

6(7, 13)(0

)

(7, 9)(0)

2(14,16)(7)

(13, 20)(0)

7(13, 20)(0)

(9,1

3)(0

)4(

16, 2

0)(7

)

(9,17)(0)8(12,20)(3)


PERT Model

Historical Evolution.Before 1957 there was no generally accepted procedure to aid the management of a project. In 1958 PERT was developed by team of engineers working on a Polaris Missile programme of the navy. This was a large project involved 250 prime contractors and about 9000 job contractors. It had about 19 million components. In such projects it is possible that a delay in the delivery of a small component might hold the progress of entire project. PERT was used successfully and the total time of completion was reduced from 7 years to 5 years.



PERT CPM













In PERT model, 3 time values are associated

with each activity. They are

i)Optimistic time = to

ii)Pessimistic time = tp

iii)Most likely time = tm

These three times provide a measure of uncertainty associated with that activity


Optimistic Time: This is the shortest possible time

in which the activity can be finished. It assumes

that every thing goes well.

Pessimistic Time: This is the longest time that an

activity could take. It assumes that every thing

goes wrong.

Most likely Time: It is the estimate of the normal

time that an activity would take. This assumes

normal delays.


Expected Time ( te):

‘te’ can be calculated by the following formula

te = (to + 4tm + tp) / 6

Example.

If a job has to = 5 days, tp = 17 days, tm = 8 days

Then Expected time for the job would be

te = (to + 4tm + tp) / 6

= (5 + 4 x 8 + 17) / 6

= 9 days


Variability of activity times

•Standard deviation and Variance are commonly used in statistics to measure the variability of number.

In PERT model, to measure the variability of an activity time duration standard deviation and variance are used.

A large standard deviation represents high variability and vice-versa.


•Calculation of Standard Deviation and Variance

Variance = (Standard deviation )2

Standard deviation =(t p – t o) / 6

•Expected length of the Critical Path = te of all the activities along the Critical Path


Probability of completing the project within a given date

Z = (TS – TE ) / σ

Where TS = Scheduled time for project completion

TE = Expected time for the project completion

σ = Standard deviation for the Network


σNetwork = √Sum of variances along the

Critical Path

= √ (σNetwork )2


Example:Construct the Network for the following project and calculate the probability of completing the project in 25 days

Activity to tm tp

1-2 2 6 10

1-3 4 8 12

2-3 2 4 6

2-4 2 3 4

3-4 0 0 0

3-5 3 6 9

4-6 6 10 14

5-6 1 3 5


1

2

3 5

4

6

2- 6 -10

4 - 8 - 12

2 - 3 - 4

2 -

4 -

6

0 - 0 - 0

6 - 4 - 10

3 - 6 - 91 - 3

- 5

1.Construction of the Network


1

2

3 5

4

6

2- 6 -10

4 - 8 - 12

2 - 3 - 4

2 -

4 -

6

0 - 0 - 0

6 - 4 - 10

3 - 6 - 91 - 3

- 5

2. Calculation of Expected time for all the activities

6

3

8

40

6

10

3

Expected Time ( te):‘te’ can be calculated by the following formula te = (to + 4tm + tp) / 6


1

2

3 5

4

6

2- 6 -10

4 - 8 - 12

2 - 3 - 4

2 -

4 -

6

0 - 0 - 0

6 - 4 - 10

3 - 6 - 91 - 3

- 5

3. Determination of Critical Path

6

3

8

40

6

10

3

Expected Duration of the project Te = 20 days

Keyto - tm - tp

te


Activity to tm tp Critical activities

σ2 = ((t p – t o) / 6)2

1-2 2 6 10 1-2 1.78

1-3 4 8 12 -

2-3 2 4 6 2-3 0.44

2-4 2 3 4 -

3-4 0 0 0 3-4 0

3-5 3 6 9 -

4-6 6 10 14 4-6 1.78

5-6 1 3 5 -

Σ σ2 = 4.00


σNetwork = √Sum of variances along the

Critical Path

= √ (σNetwork )2

= √ 4

= 2



Z = (TS – TE ) / σ




= (25 – 20) / 2

= + 2.5


From the Normal distribution Table, we get the probability of completing the project in 25 days is 99.4%


Example.The following table lists the jobs of a network along with theirtime estimates.

Activity to tm tp

1-4 3 9 27

1-3 3 6 15

1-2 6 12 30

4-5 1 4 07

3-5 3 9 27

3-6 2 5 08

5-6 6 12 30

2-6 4 19 28


a) Draw the project network.

b) What is the probability that the job will be

completed in 35 days?

c) What due date has 90% chance of being met?


3

2

1 6

4 5

6 -12 - 30

4 - 19 - 28

3 - 6 - 15 2 - 5 - 8

3 - 9 - 27

3 - 9 - 27 6 -12 - 3

0



3

2

1 6

4 5

6 -12 - 30

4 - 19 - 28

3 - 6 - 15 2 - 5 - 8

3 - 9 - 27

3 - 9 - 27 6 -12 - 3

0

2. Calculation of Expected time for all the activities

Expected Time ( te):‘te’ can be calculated by the following formula te = (to + 4tm + tp) / 6

14

18

7 5

11 14

1 - 4 - 7

4

11


3

2

1 6

4 5

6 -12 - 30

4 - 19 - 28

3 - 6 - 15 2 - 5 - 8

3 - 9 - 27

3 - 9 - 27 6 -12 - 3

0

14

18

7 5

11 14

1 - 4 - 7

4

11

3. Determination of Critical Path

Expected Duration of the project Te = 32 days

Keyto - tm - tp

te


Activity σ2 = ((t p – t o) / 6)2 σ2

1-2 ((30 – 6)/6)2 16

2-6 ((28 – 4)/6)2 16

As there are two Critical Paths, the path which gives more variance(σ2) is taken as Critical Path

Path A

Σ σ2 = 32.00


Activity σ2 = ((t p – t o) / 6)2 σ2

1-3 ((15 – 3)/6)2 4

3-5 ((27 – 3)/6)2 16

5-6 ((30 – 6)/6)2 16

Path B

Σ σ2= 36.00σ = √Σ σ2 =√ 36 = 6

Therefore the Critical Path is 1 - 3 - 5 - 6


b)


Z = (TS – TE ) / σ




= (35 – 32) / 6

= + 0.5


From the Normal distribution Table, we get the

probability of completing the project in 35 days is

69.15%


c)

The due date for 90% chance of being met.


Z = (TS – TE ) / σ

The value of Z from the table for a 90% probability is +1.28

TS = ? (to be calculated) ,TE = 32, σ = 6

i.e. 1.28 = (TS– 32) / 6

TS = 39.68 days

σ =


CPM Model

In 1958 Du Pont Company used a technique called

Critical Path Method (CPM) to schedule and control a

very large project like overhauling of a chemical plant,

there by reducing the shutdown period from 130hrs to

90 hrs saving the company 1 million dollar.

Both of these techniques are referred to as project

scheduling techniques.


PERT CPM














Cost considerations in PERT / CPM

The total cost of any project comprises of two costs.

•Direct cost - material cost, manpower loading

•Indirect cost - overheads such as managerial services, equipment rent, building rent etc.


Crash Normal

Job duration

Dir

ect

cost

Shorter the duration higher will be the Direct expenses


Job durationCrash Normal

In

dir

ect

cost

Shorter the duration lesser will be the Indirect expenses


C

ost

Job duration

NormalOptimumCrash

Indirect cost

Direct cost

Total cost


Find the lowest cost and optimum cost schedule for the following project, given the over head expenses as Rs.45/-day.

Activity Normal duration

Crashduration

Cost of crashing per day

1-2 3 1 Rs.40

2-3 4 2 Rs.40

2-4 7 3 Rs.10

3-4 5 2 Rs.20


1 2

3

4


3 -1

7 - 3

4 - 2

5 - 240

10

40

20

KeyNormal duration – Crash duration

Cost of crashing per day in Rs.


1 2

3

4

2.Determination of Critical path

3 -1

7 - 3

4 - 2

5 - 240

10

40

20

KeyNormal duration – Crash duration

Cost of crashing per day in Rs.


1 32 4

4

3 -1 4 - 2 5 - 2

7 - 3

40 40 20

1012 days

Activity crashed

Days saved

Project duration

Cost of crashing Total cost of

crashing

Over Head cost

Total cost

None 0 12 -Nil- -Nil- 45 x 12 540

3-4 2 10 20 x 2 =40 40 45 x 10 490

3-4 &2-4 1 9 20x1+10x1 =30 70 45 x 9 475

1-2 2 7 40 x 2 =80 150 45 x 7 465

2-3&2-4 2 5 40x2+10x2 =100 250 45 x 5 475

Step 1.


1 32 43 -1 4 - 2 3 - 2

7 - 3

40 40 20

1010 days

Activity crashed

Days saved

Project duration


crashing

Over Head cost

Total cost

None 0 12 -Nil- -Nil- 45 x 12 540

3-4 2 10 20 x 2 =40 40 45 x 10 490

3-4 &2-4 1 9 20x1+10x1 =30 70 45 x 9 475

1-2 2 7 40 x 2 =80 150 45 x 7 465

2-3&2-4 2 5 40x2+10x2 =100 250 45 x 5 475

Step 2.


1 32 43 -1 4 - 2 2 - 2

6 - 3

40 40 20

109 days

Activity crashed

Days saved

Project duration


crashing

Over Head cost

Total cost

None 0 12 -Nil- -Nil- 45 x 12 540

3-4 2 10 20 x 2 =40 40 45 x 10 490

3-4 &2-4 1 9 20x1+10x1 =30 70 45 x 9 475

1-2 2 7 40 x 2 =80 150 45 x 7 460

2-3&2-4 2 5 40x2+10x2 =100 250 45 x 5 475

Step 3.


1 32 41 -1 4 - 2 2 - 2

6 - 3

40 40 20

107 days

Activity crashed

Days saved

Project duration


crashing

Over Head cost

Total cost

None 0 12 -Nil- -Nil- 45 x 12 540

3-4 2 10 20 x 2 =40 40 45 x 10 490

3-4 &2-4 1 9 20x1+10x1 =30 70 45 x 9 475

1-2 2 7 40 x 2 =80 150 45 x 7 460

2-3&2-4 2 5 40x2+10x2 =100 250 45 x 5 475

Step 4.


1 32 43 -1 2 - 2 2 - 2

4 - 3

40 40 20

105 days

Activity crashed

Days saved

Project duration


crashing

Over Head cost

Total cost

None 0 12 -Nil- -Nil- 45 x 12 540

3-4 2 10 20 x 2 =40 40 45 x 10 490

3-4 &2-4 1 9 20x1+10x1 =30 70 45 x 9 475

1-2 2 7 40 x 2 =80 150 45 x 7 460

2-3&2-4 2 5 40x2+10x2 =100 250 45 x 5 475

Step 5.


PERT CPM













Documents

C.R.Krishna Prasad, BIT Bangalore-4 Operations research